This course explains how to analyze circuits that have direct current (DC) current or voltage sources. A DC source is one that is constant. Circuits with resistors, capacitors, and inductors are covered, both analytically and experimentally. Some practical applications in sensors are demonstrated.

Преподаватели

Dr. Bonnie H. Ferri

Professor

Dr. Joyelle Harris

Director of the Engineering for Social Innovation Center

Dr Mary Ann Weitnauer

Текст видео

The topic of this problem is Current Division. In this problem, we have a current source and we have three resistors, in series and parallel. And we want to find the current I1 through the 60 kilo ohm resistor. We want to find the current I2 through the 40 kilo ohm at the 80 kilo ohm resistance, the same current flows for both of those. And we want to find the output voltage V out across the 80 kilo ohm resistor. The V out is measured plus to minus top to bottom across that resistor. So in order to do this, we going to use Current Division and we'll use the well-known equation for Current Division. If you're not familiar with the derivation of current division, how you would divide the current between those resistors, then I would refer you back to a previous video on Current Division that shows how Current Division is actually derived for a single node pair circuit which is essentially what we have here if we combine the 80k and 40k into a single resistor. So here we go. The first thing we want to do is we want to find I1. I1 is going to be equal to the current source which is 0.9 milliamps. And this 0.9 milliamps is split between the route through to the 60 kilo ohm resistor and the 40 kilo ohm resistor. So to find out how much is flowing through the 60 kilo ohm resistor, we take the resistance of the opposite path, which is a 40 kilo ohm resistor plus an 80 kilo ohm resistor. And we divide it through by the total sum of the resistances through the two paths, so it's 40K plus 80K for the rightmost path plus the other path which is our 60 kilo ohm resistor. And if we solve for I1, in this problem, then we end up with I1 equal to 0.6 milliamps. Similarly, if we look at I2, I2 is going to be equal to the current source which is 0.9 milliamps times the resistor associated with the opposite leg which is the 60 kilo ohm resistor, again divided by the sum of the 40k plus 80k and the 60k resistor. And if we solve for that, we end up with 0.3 milliamps. So we see that, first of all, that the largest amount of the current flows through the path of least resistance which is the 60 kilo ohm resistor, and the least amount of current flows through the other path which is the 120 kilo ohm resistor. So in addition to that, we can also see that the sum of I1 plus I2 as a check is equal to the current source 0.9 milliamps, so we indeed did the problem correct. Next, we look for the output voltage, V out. We know that V out is going to be equal to I2 times 80 kilo ohms. And we know what I2 is. I2 is 0.3 milliamps. If we multiply that by the 80k of resistance then we end up with the V out which is equal to 24 volts. So our volt is drop across the 80 kilo ohm resistor is 24 volts.